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September 30th, 2015, 11:06 PM   #1
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height of prime ideal

can someone please help me to show this result:
Let A be a factoriel ring,and P included in A be a prime ideal
show this equivalence:
a)p is a principal ideal
b) ht(p) ≤1
thanks in advance
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October 11th, 2015, 11:56 AM   #2
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Suppose that $P$ is a principal prime ideal generated by $a$, let $Q$ be a prime strictly contained in $P$ and $b$ an element of $Q$. We can write $b =ca$. Since A is factoriel, $c = p_1...p_n$ where $p_i$ is irredcible. We deduce that $b = p_1...p_na$. Thus, $p_1(p_2...p_na)$ is an element of $Q$. Since $Q$ is a prime, $p_1\in Q$ or $p_2...p_nc \in Q$.
Suppose that $p_1\in Q$ since $Q$ is contained in $P$, we can write $p_1 = da$ we deduce that $d$ is invertible since $p_1$ is irreducible.This implies that $a\in Q$ and henceforth $P = Q$.This is a contradiction. Thus $p_2(p_3..p_n)a \in Q$. We repeat the previous step $n-1$ times to deduce that $a\in Q$. This is a contradiction thus the height of $P$ is 1.

On the other hand, suppose that the height of the prime ideal $P$ is 1. Let $a$ be a non zero element of $P$, we can write $a = p_1...p_n$ where $p_i$ is irreducible since $A$ is factoriel, suppose that $p_i$ is not in $P$ for every $i$, this implies that $p_1p_2$ is not in $P$ since $P$ is a prime. For the same reason, $p_1p_2p_3,....,p_1...p_n =a$ is not in $P$ a fact which is a contradiction. Thus $P$ contains an irreducible element that we call $a$ Since $(a)$ the ideal generated by $a$ is a prime (since $A$ is factoriel and $a$ irreducible) and the height of $P$ is 1, we deduce that $(a) = P$.

Last edited by Aristide Tsemo; October 11th, 2015 at 12:00 PM.
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